A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass 3.8 kg, is moving upward at 19 m/s and the other ball, of mass 1.6 kg, is moving downward at 11 m/s. How high do the combined two balls of putty rise above the collision point? (Neglect air drag.)

Explanation:

The inelastic collision is modelled by using the Principle of Momentum Conservation:

The final velocity is:

The maximum height of the composite system is:

5.21m

Explanation:

We are given that;

mass of first ball; m1 = 3.8kg

Speed of first ball; v1 = 19 m/s upwards

Mass of second ball; v2 = 1.6 kg

Speed of second ball; v2 = 11 m/s downwards

From conservation of linear momentum,

m1v1 + m2v2 = m_t•v_t

Where,

m1v1 is momentum of first ball

m2v2 is momentum of second ball

m_t•v_t is the combined momentum of the 2 balls.

Let's make v_t the subject

v_t = [m1v1 + m2v2]/m_t

m_t is the combined mass of both balls.

Since first and second ball are moving in opposite directions, thus;

v_t = [m1v1 - m2v2]/m_t

Thus, m_t = 3.8 + 1.6 = 5.4kg

Thus, plugging in the relevant values, we have;

v_t = [(3.8 x 19) - (1.6 x 11)]/5.4

v_t = [(3.8 x 19) - (1.6 x 11)]/5.4

v_t = (72.2 - 17.6)/5.4

v_t = 54.6/5.4 = 10.11 m/s

Now, from equation of motion,

v² = u² + 2gh

Where u is initial velocity which is now v_t while final velocity v is zero.

Now, since gravity is acting against motion, g = - 9.8m/s²

Thus,

v² = u² + 2gh gives;

0² = 10.11² - (2 x 9.8 x h)

19.6h = 102.212

h = 102.212/19.6

h = 5.21 m

answer; /// the above statement is (true);

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

don't remove my answer im trying my best to answer them

explanation: